Taylor series of sin(x) needs to be modified for cos(x) in a better way, please explain how to do this

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John Cleary 2022-4-28

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回答： Akshat Dalal 2023-9-14

    function [approx] = T_C(angle,counter)
approx=0;
radianangle= angle*pi/180;
polynum=1;
polynum=radianangle;
approx=approx+polynum;
while abs(counter/polynum)<1
    n=n+1;
    polynum=polynum*-1*anglerad*2/(2*n+1)/(2*n);
    approx=approx+polynum;
end
end

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Jan 2022-4-28

Please elaborate the question. What doesd "in a better way" mean?

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Answer 1

Akshat Dalal 2023-9-14

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Hi John,

I understand you want to find the value of sin(x) using Taylor series expansion. I believe your code should be modified as follows:

polynum=polynum*-1*(anglerad^2)/(2*(n+1))/(2*n);

You could modify the code accordingly for cos(x). You can find the both the Taylor series expansions here: https://en.wikipedia.org/wiki/Taylor_series